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JEE Main 20154 April
Exam
JEE Main
Year
2015
Date
4 April
Subject
Mathematics
Chapter
Parabola

JEE Main 2015 · 4 April · Mathematics · Parabola

Let O be the vertex and Q be any point on the parabola, x ^ 2 = 8y...

Let OO be the vertex and QQ be any point on the parabola, x2=8y{x}^{2} = 8y. If the point PP divides the line segment OQOQ internally in the ratio 1:31: 3, then the locus of PP is

Options

  1. Ax2=2y{x}^{2} = 2y
  2. Bx2=y{x}^{2} = y
  3. Cy2=x{y}^{2} = x
  4. Dy2=2x{y}^{2} = 2x

Correct answer

A. x2=2y{x}^{2} = 2y

Step-by-step solution

General point on x2=8y{x}^{2} = 8y is Q(4t,2t2)Q\left(4t, 2{t}^{2}\right)

Let P(h,k)P\left(h, k\right) divide OQOQ in ratio 1:31: 3

\left(h, k\right) = \left(\dfrac{1\left(4t\right) + 3\left(0\right)}{1 + 3}, \dfrac{1\left(2{t}^{2}\right) + 3\left(0\right)}{1 + 3}\right)</p><p>(h,k)=(t,2t24)</p><p>\left(h , k\right) = \left(t,\dfrac{2{t}^{2}}{4}\right)

h=t and k=t22h = t \text{ and } k = \dfrac{{t}^{2}}{2}

k=h22k = \dfrac{{h}^{2}}{2}

⇒x2=2y⇒ {x}^{2} = 2y is the required locus.

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